SGU196 水题。答案直接就是每个点的度数平方和。

#### 196. Matrix Multiplication

time limit per test: 0.25 sec. memory limit per test: 65536 KB

input: standard output: standard

Let us consider an undirected graph G = <V, E> which has N vertices and M edges. Incidence matrix of this graph is an N × M matrix A = {aij}, such that aij is 1 if i-th vertex is one of the ends of j-th edge and 0 in the other case. Your task is to find the sum of all elements of the matrix ATA where AT is A transposed, i.e. an M × N matrix obtained from A by turning its columns to rows and vice versa.

Input

The first line of the input file contains two integer numbers — N and M (2 le N le 10,000, 1 le M le 100,000). 2M integer numbers follow, forming M pairs, each pair describes one edge of the graph. All edges are different and there are no loops (i.e. edge ends are distinct).

Output

Output the only number — the sum requested.

Sample test(s)

Input

4 4 1 2 1 3 2 3 2 4

Output

18

/* ***
Author :kuangbin
Created Time :2014/11/21 22:57:16
File Name :E:\2014ACM\SGU\SGU196.cpp
************************************************ */

#include <stdio.h>

#include <string.h>

#include

#include

#include

#include

#include

#include

#include

#include <math.h>

#include <stdlib.h>

#include <time.h>
using namespace std;
int du[10010];

int main()
{
//freopen(“in.txt”,”r”,stdin);
//freopen(“out.txt”,”w”,stdout);
int n,m;
while(scanf(“%d%d”,&n,&m) == 2){
memset(du,0,sizeof(du));
int u,v;
while(m–){
scanf(“%d%d”,&u,&v);
du[u]++;
du[v]++;
}
int ans = 0;
for(int i = 1;i <= n;i++)
ans += du[i]*du[i];
printf(“%d\n”,ans);
}
return 0;
}

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